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Can someone explain to me intuitively and without much technical stuff the following:

A hermitian line bundle is a complex line bundle with a hermitian metric. I think of this as a bundle over my manifold $M$ that assigns to each point in $M$ a complex plane. A section is then a function that takes a point the manifold and it sends it to a complex number in the associated complex plane. Therefore a section looks like a complex function on the manifold. However I was told that this is only true locally in general. If the line bundle is trivial then it is true globally but it is possible depending on the line bundle for this intuition to break down. Can someone explain why to me and perhaps give me an example? If my intuition so far is wrong can you explain how and why?

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$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Cpx}{\mathbf{C}}\newcommand{\Proj}{\mathbf{P}}$If you're seeking intuition about how a line bundle can be non-trivial, the simplest example is the Möbius strip viewed as the total space of a real line bundle over a circle.

In detail, view $\Reals^{2} \to \Reals$ as the trivial real line bundle via projection to the first factor. Divide $\Reals^{2}$ by the glide reflection $\gamma(x, v) = (x + 1, -v)$, and view the result, by projection to the first factor, as a line bundle over the circle $\Reals/\mathbf{Z}$. A fundamental domain for the action is the strip $[0, 1] \times \Reals$, and $(0, v) \sim (1, -v)$; that is, the edges of the strip are glued with a half-twist. It's easy (intermediate value theorem) to show this line bundle has no continuous, non-vanishing section, so the bundle is non-trivial.


Your intuition about sections of Hermitian line bundles being local complex-valued functions is correct.

The simplest Hermitian line bundles (over complex curves) are harder to visualize than real line bundles because their total spaces are complex surfaces, hence real four-manifolds. Take $M = \Cpx\Proj^{1}$, the complex projective line (real two-sphere). Removing many details:

  • In the trivial complex line bundle, the set of unit vectors is (diffeomorphic to) $S^{2} \times S^{1}$.

  • In the tangent bundle $TM = TS^{2}$, the set of unit vectors is (diffeomorphic to) the special orthogonal group $SO(3)$: A point $p$ of $S^{2}$ is a unit vector in $\Reals^{3}$, a unit tangent vector $v$ at $p$ is an element of $\Reals^{3}$ orthogonal to $p$, and the ordered triple $(p, v, p \times v)$ may be viewed as an element of $SO(3)$. This association is clearly bijective (because reading the columns from an element of $SO(3)$ recovers a point of the sphere and a unit tangent vector, and these uniquely determine the third column) and smooth (write either map in coordinates coming from the entries of $3 \times 3$ real matrices).

  • A possibly less familiar example is the complement $L$ of a point $q$ in the complex projective plane $\Cpx\Proj^{2}$, which is the total space of a non-trivial Hermitian line bundle in which the set of unit vectors is (diffeomorphic to) $S^{3}$. (Pick a complex projective line $M$ not containing $q$. For each point $x$ of $L$, let $\ell$ be the unique line through $q$ and $x$, and let $p = M \cap \ell$. The mapping $x \mapsto p$ is a line bundle projection. The bundle of unit vectors is (essentially) a sphere in $\Cpx\Proj^{2}$ centered at $q$.)

The preceding examples are distinct because their sets of unit vectors are mutually non-diffeomorphic. There are infinitely many Hermitian lines bundles over the projective line, classified by an integer, the Chern number. It turns out the preceding examples have Chern number $0$, $2$, and $1$ respectively.

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